**4.B] State Heisenberg uncertainty Principle. Show that electron does not exists inside the nucleus by this Principle.**

**Answer:-**

**Statement :-**

The simultaneous determination of the exact position and momentum of a moving particle is impossible.

**Explanation :**–

According to this principle if \Delta x is the error involved in the measurement of position and \Delta p_{x} is the error involved in the measurement of momentum during their simultaneous measurement, then the product of the corresponding uncertainties is given by

\Delta x\Delta p_{x}\ge \frac{h}{4\Pi} \Delta E\Delta r \ge \frac{h}{4\Pi} \Delta\Theta\Delta L\ge \frac{h}{4\Pi}The product of the errors is of the order of Planck’s constant. If one quantity is measured with high accuracy then the simultaneous measurement of the other quantity becomes less accurate.

**Non-exixtance electrons inside the nucleus **: Beta rays are emitted by the nucleus. When it was first observed it was believed that electrons exist inside the nucleus and are emitted at certain instant. If the electron can exist inside the atomic nucleus then uncertainty in its position must not exceed the diameter of the nucleus. The diameter of the nucleus is of the order of \Delta x_{max} is 10^{-14}m . Applying Heisenberg’s uncertainty principle for an electron expected to be inside the nucleus we get

therefor, the electron should possess momentum

p_{min}\simeq p_{min}=5.276\times 10^{-21}kgms^{-1}Relativistically the energy of the electron us given by the equation

E=\sqrt{p^{2}c^{2}+m_{0}^{2}c^{4}}Since, m_{0}^{2}c^{4}, we get E=pc. therefore

E_{min}=p_{min}c=5.276\times 10^{-21}\times 3\times 10^{8} E_{min}=1.58\times 10^{-12}J E_{min}=\frac{1.58\times 10^{-12}}{1.6\times 10^{-19}}=9.9MevConclusion : According to experiments, the energy associated with the beta ray (electron) emission is around 3 MeV which is much lesser than the energy of the electron expected to be inside the nucleus 9.9 MeV. Hence electrons do not exist inside the nucleus..

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